Introduction Vibrating String Math 531 - Partial Differential Equations Vibrating String Joseph M. Mahaffy, [email protected] Department of Mathematics and Statistics Dynamical Systems Group Computational Sciences Research Center San Diego State University San Diego, CA 92182-7720 http://jmahaffy.sdsu.edu Spring 2020 Joseph M. Mahaffy, [email protected] Vibrating String | (1/14) Introduction Vibrating String Outline 1 Introduction Derivation String Equation 2 Vibrating String Physical Interpretation Traveling Wave Joseph M. Mahaffy, [email protected] Vibrating String | (2/14) Introduction Derivation Vibrating String String Equation Introduction An important application of PDEs is the investigation of vibrations of perfectly elastic strings and membranes Perturbed String v (x, y) u Equilibrium α (highly stretched) Consider a particle at position α in a highly stretched string Assume a small displacement as seen above Joseph M. Mahaffy, [email protected] Vibrating String | (3/14) Introduction Derivation Vibrating String String Equation Derivation 1 Simplify by assuming the displacement is only vertical, y = u(x; t) T(x + ∆x, t) θ(x + ∆x, t) θ(x, t) T(x, t) x x + ∆x Apply Newton's Law to an infinitesimally small segment of string between x and x + ∆x Assume string has mass density ρ0(x), so mass is ρ0(x)∆x Joseph M. Mahaffy, [email protected] Vibrating String | (4/14) Introduction Derivation Vibrating String String Equation Derivation 2 Newton's Law acting on string considers all forces T(x + ∆x, t) θ(x + ∆x, t) Forces include gravity, resistance, and tension - \body" forces Assume string is perfectly flexible, so no bending resistance θ(x, t) T(x, t) This implies primary force is tangent to the string at all points x x + ∆x Tension is the tangential force with dy @u = = tan(θ(x; t)) dx @x Joseph M. Mahaffy, [email protected] Vibrating String | (5/14) Introduction Derivation Vibrating String String Equation Derivation 3 Newton's Law gives F~ = m~a, which is @2u ρ (x)∆x = T (x + ∆x; t) sin(θ(x + ∆x; t)) 0 @t2 −T (x; t) sin(θ(x; t)) + ρ0(x)∆xQ(ξ; t); where ξ 2 [x; x + ∆x] and Q(ξ; t) are any \body" accelerations, such as gravity or air resistance. Dividing by ∆x and taking the limit as ∆x ! 0 gives @2u @ ρ (x) = T (x; t) sin(θ(x; t)) + ρ (x)Q(x; t): 0 @t2 @x 0 For θ \small," let @u sin(θ) = tan(θ) = ≈ sin(θ) @x cos(θ) Joseph M. Mahaffy, [email protected] Vibrating String | (6/14) Introduction Derivation Vibrating String String Equation String Equation From previous results, obtain String Equation @2u @ @u ρ (x) = T (x; t) + ρ (x)Q(x; t): 0 @t2 @x @x 0 If the string is perfectly elastic, then T (x; t) ≈ T0 constant, which is equivalent to almost uniform stretching along string @2u @2u ρ (x) = T + ρ (x)Q(x; t): 0 @t2 0 @x2 0 If the body force is small and density is constant, then @2u @2u = c2 ; @t2 @x2 where c2 = T0 . ρ0 Joseph M. Mahaffy, [email protected] Vibrating String | (7/14) Introduction Physical Interpretation Vibrating String Traveling Wave Vibrating String - Separation of Variables The vibrating string satisfies the following: @2u @2u PDE: = c2 , BC: u(0; t) = 0, @t2 @x2 u(L; t) = 0. IC: u(x; 0) = f(x), ut(x; 0) = g(x). This vibrating string problem or wave equation has fixed ends at x = 0 and x = L and initial position, f(x), and initial velocity, g(x). As before, we apply our separation of variables technique: u(x; t) = φ(x)h(t); so h00 φ00 φ00h = c2φh00 or = = −λ. c2h φ Joseph M. Mahaffy, [email protected] Vibrating String | (8/14) Introduction Physical Interpretation Vibrating String Traveling Wave Vibrating String - SL Problem The homogeneous BCs give: φ(0) = 0 and φ(L) = 0: The Sturm-Liouville Problem becomes φ00 + λφ = 0 with φ(0) = 0 = φ(L): As before, we saw λ ≤ 0 results in the trivial solution. If we take λ = α2 > 0, then φ(x) = c1 cos(αx) + c2 sin(αx); nπ where the BCs show c1 = 0 and α = L for nontrivial solutions. The eigenvalues and associated eigenfunctions are n2π2 nπx λ = with φ (x) = sin : n L2 n L Joseph M. Mahaffy, [email protected] Vibrating String | (9/14) Introduction Physical Interpretation Vibrating String Traveling Wave Vibrating String - Superposition The other second order DE becomes: n2π2 h00 + c2h = 0; L2 which has the solution nπct nπct hn(t) = c1 cos L + c2 sin L : It follows that nπct nπct nπx un(x; t) = An cos L + Bn sin L sin L The Superposition principle gives: 1 X nπct nπct nπx u(x; t) = An cos L + Bn sin L sin L n=1 Joseph M. Mahaffy, [email protected] Vibrating String | (10/14) Introduction Physical Interpretation Vibrating String Traveling Wave Vibrating String - ICs The initial position gives: 1 X nπx u(x; 0) = f(x) = An sin L ; n=1 where Z L 2 nπx An = f(x) sin L dx: L 0 The velocity satisfies 1 X nπct nπct nπc nπx ut(x; t) = −An sin L + Bn cos L L sin L : n=1 The initial velocity gives: 1 X nπc nπx ut(x; 0) = g(x) = Bn L sin L ; n=1 where Z L 2 nπx Bn = g(x) sin L dx: nπc 0 Joseph M. Mahaffy, [email protected] Vibrating String | (11/14) Introduction Physical Interpretation Vibrating String Traveling Wave Physical Interpretation Physical Interpretation: Model for vibrating string 1 X nπct nπct nπx u(x; t) = An cos L + Bn sin L sin L n=1 Musical instruments Each value of n gives a normal mode of vibration Intensity depends on the amplitude p 2 2 An An cos(!t)+Bn sin(!t) = A + B sin(!t+θ); θ = arctan n n Bn Time dependence is simple harmonic with circular nπc frequency, L , which is the number of oscillations in 2π units of time The sound produced consists of superposition of the infinite number of natural frequencies, n = 1; 2; ::: Joseph M. Mahaffy, [email protected] Vibrating String | (12/14) Introduction Physical Interpretation Vibrating String Traveling Wave Physical Interpretation Physical Interpretation (cont): The normal mode, n = 1, is called the first harmonic or fundamental mode πc This mode has circular frequency, L Higher natural frequencies have higher pitch q Fundamental frequency varied by changing, c = T0 ρ0 Tune by changing tension, T0 Different ρ0 for different strings (range of notes) Musician varies pitch by varying the length L (clamping string) Higher harmonics for stringed instruments are all integral multiples (pleasing to the ear) Joseph M. Mahaffy, [email protected] Vibrating String | (13/14) Introduction Physical Interpretation Vibrating String Traveling Wave Traveling Wave Traveling Wave: Show that the solution to the vibrating string decomposes into two waves traveling in opposite directions. At each t, each mode looks like a simple oscillation in x, which is a standing wave The amplitude simply varies in time The standing wave satisfies: nπx nπct 1 nπ 1 nπ sin L sin L = 2 cos L (x − ct) − 2 cos L (x + ct) 1 nπ 2 cos L (x − ct) produces a traveling wave to the right with velocity c 1 nπ 2 cos L (x + ct) produces a traveling wave to the left with velocity −c By superposition (later d'Alembert's solution) u(x; t) = R(x − ct) + S(x + ct) Joseph M. Mahaffy, [email protected] Vibrating String | (14/14).